What is the integral from 0 to 1-x of 1 ?

The integral from 0 to 1-x of 1 is -x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1-x of 1

\int_{0}^{1 - x} 1 d y

- x + 1

Solution steps

\int_{0}^{1 - x} 1 d y

Integral of a constant:

\int a d x = a x

= [1 \cdot y]_{0}^{1 - x}

Simplify

= [y]_{0}^{1 - x}

Compute the boundaries:

- x + 1

= - x + 1

\int_{0}^{1} \frac{x}{x ^{2} - x + 1} d x

\int_{0}^{a} (a x^{2} - x^{3}) d x

\int_{\frac{1}{2}}^{1} (x^{- 3} - 7) d x

\int_{0}^{6} \frac{1}{x ^{0.8}} d x

\int_{1}^{e} \frac{( 1 + ln ( x ) )}{x} d x

What is the integral from 0 to 1-x of 1 ?
The integral from 0 to 1-x of 1 is -x+1